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Fermi's golden rule for transitions between single-particle states $a$ and $b$ is $$ \Gamma_{ a \to b} = \frac{2\pi}{\hbar}\vert M_{ab} \vert^2\delta(\epsilon_a - \epsilon_b) \, .\tag{1} $$ Here $\epsilon_i$ is the single-particle energy of state $i$. In fermionic systems, we often augment this by occupation factors: the initial state needs to be filled and the final state empty, and therefore we multiply the above by $$ f(\epsilon_a) (1-f(\epsilon_b)) $$ by hand where $f$ is the Fermi occupation function. Can this be justified in a formal way?

I have tried starting from (1), assuming that the matrix element is for many particle (Fock) states. Then, assuming that the perturbation is single-particle in nature and with initial state given by the Fermi sea, I get something like $$ \langle \Psi \vert\ \sum_{\alpha\beta} V_{\alpha\beta}\ a^{\dagger}_{\alpha}a_{\beta}\ \vert \textrm{FS} \rangle = \cases{ V_{\gamma\delta} \textrm{ if } \vert \Psi\rangle = a^{\dagger}_{\gamma}a_{\delta} \vert \textrm{FS} \rangle \textrm{ and } \epsilon_{\gamma} > 0,\ \epsilon_{\delta} < 0 \\ 0 \textrm{ else } } , $$ for some $\gamma,\delta$, where $V_{\gamma\delta}$ is the single-particle matrix element. But I don't really know how to continue from here. In particular, I don't know how to treat finite temperature.

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You're on the right track with assuming the perturbation is single-particle in nature. We can write the initial state in the occupation number representation as $$ |i\rangle = |f_a, f_b\rangle . $$ where $f_a$ and $f_b$ are some arbitrary, as-yet unspecified occupations. After a single-particle transition $$ |a\rangle \longrightarrow |b \rangle, $$ the final state must necessarily become $$ |f \rangle = |f_a - 1, f_b + 1 \rangle. $$ Therefore, the actual matrix element we should be inserting into the Fermi golden rule is

\begin{align} \langle f | V | i \rangle &= \langle f_a - 1, f_b + 1 | V | f_a, f_b \rangle \\ &= \frac{1}{\sqrt{f_a}\sqrt{1 - f_b}}\langle f_a, f_b |c_b c^\dagger_a V | f_a, f_b \rangle. \end{align} NOTE: The normalisation factor for fermionic operators is different to the bosonic case.

If I take your second-quantised operator $$ V = \sum_{i j} V_{ij} c_i^\dagger c_j, $$ then only the $V_{ab}$ term gets pulled out of the sum, and you can check the following using the commutation relations: $$ \langle f_a, f_b |c_b c^\dagger_a c_b^\dagger c_a | f_a, f_b \rangle = -f_a(1 - f_b). $$

Therefore, Fermi's golden rule becomes $$ \Gamma_{i\rightarrow f} = \frac{2 \pi}{\hbar} |V_{ab}|^2 f_a (1- f_b) \delta(\epsilon_a - \epsilon_b),$$ where the square roots above partially cancelled the occupation factors from the expectation value, and squaring the matrix element cancels the minus sign.

The generalisation to finite temperature follows immediately by promoting the occupations $f_a$ and $f_b$ to be functions of temperature, eg the Fermi-Dirac distribution.

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